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The centered second difference differs from the second derivative by at most $M a^{2} / 12$ when $∣ u^{''''} ∣ \leq M$

Statement

Let $J \subseteq R$ be a nonempty open interval (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), let $u : J \to R$ be of class $C^{4}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), let $x \in J$ and $a > 0$ satisfy $[x - a, x + a] \subseteq J$ , and put

$M : = \sup_{s \in [x - a, x + a]} ∣ u^{''''} (s) ∣ < + \infty .$

Then

$∣ \frac{u (x + a) - 2 u (x) + u (x - a)}{a^{2}} - u^{''} (x) ∣ \leq \frac{M a^{2}}{12} .$

Facts & Assumptions

Given: The open interval $J$ , the $C^{4}$ function $u$ , the point $x \in J$ and $a > 0$ with $[x - a, x + a] \subseteq J$ , and the bound $M$ .

[F1]

Fundamental theorem of calculus: for a $C^{1}$ function $φ$ on an interval and points $r < s$ in it, $\int_{r}^{s} φ^{'} = φ (s) - φ (r)$ ; moreover $\int_{s}^{r} = - \int_{r}^{s}$ , and the integral is additive over subintervals (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

[F2]

The bound $M$ is the maximum of the continuous function $∣ u^{''''} ∣$ on the compact set $[x - a, x + a]$ , so $∣ u^{''''} ∣ \leq M$ there (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, Open ball, closed ball and sphere in a metric space); derivatives and sums compose by the chain and algebra rules (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[F3]

(Second-order integral remainder.) If $φ \in C^{2}$ on an interval containing $[0, a]$ , then $φ (a) = φ (0) + φ^{'} (0) a + \int_{0}^{a} (a - s) φ^{''} (s) d s$ . By the product rule, $\frac{d}{d s} ((a - s) φ^{'} (s)) = - φ^{'} (s) + (a - s) φ^{''} (s)$ ; integrating this identity with [F1] and rearranging gives the displayed formula.

Proof

technique · direct; apply the second-order integral remainder to two affine scans of $u$ and bound the double remainder

1.1givenF1F2F3algebra

Fix $s \in [0, a]$ and apply [F3] to $φ_{+} (r) : = u (x + r)$ on $[0, s]$ : since $φ_{+}^{''} (r) = u^{''} (x + r)$ and $φ_{+}^{'} (0) = u^{'} (x)$ , $u (x + s) = u (x) + u^{'} (x) s + \int_{0}^{s} (s - r) u^{''} (x + r) d r .$ Applying [F3] to $φ_{-} (r) : = u (x - r)$ on $[0, s]$ , with $φ_{-}^{''} (r) = u^{''} (x - r)$ , $u (x - s) = u (x) - u^{'} (x) s + \int_{0}^{s} (s - r) u^{''} (x - r) d r .$

2.1step 1.1F1F3algebra

Add the two identities of step 1.1 with $s = a$ and subtract $2 u (x)$ : $u (x + a) - 2 u (x) + u (x - a) = \int_{0}^{a} (a - s) (u^{''} (x + s) + u^{''} (x - s)) d s .$ Apply [F3] once more to the $C^{2}$ function $r \mapsto ψ (r) : = u^{''} (x + r) + u^{''} (x - r)$ on $[0, s]$ , whose value and first derivative at $0$ are $ψ (0) = 2 u^{''} (x)$ and $ψ^{'} (0) = 0$ : $u^{''} (x + s) + u^{''} (x - s) = 2 u^{''} (x) + \int_{0}^{s} (s - r) (ψ^{''} (r)) d r = 2 u^{''} (x) + \int_{0}^{s} (s - r) (u^{''''} (x + r) + u^{''''} (x - r)) d r .$

3.1step 2.1F1F2algebra∎

Substitute step 2.1 into step 2.1's first identity to obtain $u (x + a) - 2 u (x) + u (x - a) - a^{2} u^{''} (x) = \int_{0}^{a} (a - s) \int_{0}^{s} (s - r) (u^{''''} (x + r) + u^{''''} (x - r)) d r d s .$ Bounding the inner integral by $2 M s^{2} / 2 = M s^{2}$ via [F2] and [F1] (monotone bounds for the Riemann integral), and then $\int_{0}^{a} (a - s) M s^{2} d s = M a^{4} / 12$ , gives $∣ u (x + a) - 2 u (x) + u (x - a) - a^{2} u^{''} (x) ∣ \leq \frac{M a^{4}}{12} .$ Dividing by $a^{2} > 0$ proves the statement.

Remarks

The constant $1 / 12$ is sharp for the class $C^{4}$ with a uniform fourth-derivative bound: the estimate is an equality for a quartic profile $u (x) = x^{4} / 24$ with $u^{''''} = 1$ after comparing the two required bounds, and the factor arises solely from $\int_{0}^{a} (a - s) s^{2} d s$ . The lemma states only the upper bound.
Only derivatives up to order four on the closed interval $[x - a, x + a]$ are used; the function need not be periodic and no boundary conditions enter.

The centered second difference differs from the second derivative by at most $M a^{2} / 12$ when $∣ u^{''''} ∣ \leq M$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The centered second difference differs from the second derivative by at most Ma2/12Ma2/12 when ∣u′′′′∣≤M∣u′′′′∣≤M

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The centered second difference differs from the second derivative by at most $M a^{2} / 12$ when $∣ u^{''''} ∣ \leq M$